A light ray emerging from the point source placed at `P(1,3)` is reflected at a point Q in the axis of x. If the reflected ray passes through the point `R(6, 7)`, then the abscissa of Q is:

Options

A. ` (5)/(2)`
B. 3
C. `(7)/(2)`
D. 1

Answer

To solve the problem step by step, we need to find the abscissa of point Q where the light ray reflects off the x-axis. Here’s how we can do it:

### Step 1: Define the Points
Let the coordinates of point Q on the x-axis be \( Q(x, 0) \). The point P from which the light ray originates is \( P(1, 3) \) and the point R through which the reflected ray passes is \( R(6, 7) \).

### Step 2: Use the Law of Reflection
According to the law of reflection, the angle of incidence is equal to the angle of reflection. This means that the tangent of the angles formed by the lines from P to Q and from Q to R will be equal.

### Step 3: Calculate the Tangents
1. **Finding \( \tan(\alpha) \)** (angle of incidence):
- The vertical distance (perpendicular) from P to the x-axis is \( 3 - 0 = 3 \).
- The horizontal distance (base) from P to Q is \( x - 1 \).
- Therefore,
\[ \tan(\alpha) = \frac{3}{x - 1} \]

2. **Finding \( \tan(\beta) \)** (angle of reflection):
- The vertical distance (perpendicular) from Q to R is \( 7 - 0 = 7 \).
- The horizontal distance (base) from Q to R is \( 6 - x \).
- Therefore,
\[ \tan(\beta) = \frac{7}{6 - x} \]

### Step 4: Set the Tangents Equal
Since \( \tan(\alpha) = \tan(\beta) \), we can set the two expressions equal to each other:
\[ \frac{3}{x - 1} = \frac{7}{6 - x} \]

### Step 5: Cross-Multiply
Cross-multiplying gives us:
\[ 3(6 - x) = 7(x - 1) \]

### Step 6: Expand and Simplify
Expanding both sides:
\[ 18 - 3x = 7x - 7 \]
Now, rearranging the equation:
\[ 18 + 7 = 7x + 3x \]
\[ 25 = 10x \]

### Step 7: Solve for x
Dividing both sides by 10:
\[ x = \frac{25}{10} = \frac{5}{2} \]

### Conclusion
The abscissa of point Q is \( \frac{5}{2} \).